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Weride·Software Engineer·Technical Phone Screen·Senior

Senior
Apr 2026

Summary

WeRide software engineer interview with a pretty gnarly string parsing problem split into two parts. The algorithmic depth they expected was no joke, and the O(1) space constraint on part one really threw me.

Questions Asked (2)

Q1

Given a string expression with lowercase variables, '+', '*', '(', and ')', fully distribute all multiplications over additions and return the equivalent expression using only '+'. In part one, assume '+' only appears inside parentheses, and implement an O(n) time solution using O(1) extra space beyond the output without using a stack.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The O(1) space constraint is where I got stuck.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and walk through a simple example to demonstrate the distribution process. Then, propose a single-pass algorithm that processes the string character by character, maintaining the current term and accumulated result, leveraging the fact that '+' only appears inside parentheses to avoid needing a stack. Finally, discuss how to achieve O(n) time and O(1) extra space by carefully managing state variables.

Pro tip: Emphasize that the restriction of '+' only inside parentheses allows a stackless solution; explicitly state that you are exploiting this constraint to avoid the typical stack-based approach for nested expressions.

1. Clarify constraints and edge cases

Confirm that variables are single lowercase letters, '+' only appears inside parentheses, and the expression is fully parenthesized. Discuss edge cases like empty string, single variable, or nested parentheses.

2. Walk through a simple example

Take a small expression like 'a*(b+c)' and manually distribute to show the expected output 'a*b + a*c'. This demonstrates understanding and sets the stage for the algorithm.

3. Design the algorithm

Propose a single left-to-right scan. Maintain a current term (string) and a result (string). When encountering '(', start a new term; when encountering ')', append the current term to the result; when encountering '*', multiply the current term by the next variable; when encountering a variable, append it to the current term. Since '+' only appears inside parentheses, we can handle it by splitting the current term into multiple terms and appending each to the result.

4. Analyze complexity and space

Explain that each character is processed once, so time is O(n). Extra space is O(1) beyond the output because we only use a few pointers and temporary strings that are part of the output.

5. Discuss trade-offs and alternatives

Mention that a stack-based approach would be more general but uses O(n) space. Highlight that the stackless solution is possible due to the problem's constraints, showing awareness of trade-offs.

Key Points to Mention

  • The constraint that '+' only appears inside parentheses enables a stackless solution.
  • Single-pass O(n) time complexity with O(1) extra space beyond the output.
  • Handling of nested parentheses without a stack by maintaining state variables.
  • The algorithm must correctly distribute multiplication over addition, e.g., a*(b+c) -> a*b + a*c.
  • Edge cases: empty string, single variable, deeply nested parentheses.
  • Comparison with the general stack-based approach and why it's not needed here.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Q2

Extend the solution to handle the general case where '+' can appear outside parentheses as well, including nested parentheses. Use a stack to track current multiplication segments, targeting O(n) time and O(n) space, and analyze the complexity and key edge cases.

Algorithms & Data StructuresSystem Design
Author's notes

This part felt more natural to me since using a stack for nested structure is pretty standard.

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AI HintsAI Generated

Suggested Approach

Clarify the problem: evaluate an expression with + and * operators, parentheses, and possibly nested parentheses, using a stack to handle multiplication segments. Outline a two-pass or single-pass stack-based algorithm that maintains a current segment and a result, pushing/popping on parentheses. Then analyze time and space complexity and discuss edge cases like negative numbers, multiple digits, and empty input.

Pro tip: Emphasize that the stack stores intermediate results and signs, and that multiplication has higher precedence, so you process it immediately while deferring addition. Mention that this approach naturally extends to nested parentheses by saving the current state before entering a parenthesis and restoring it after.

1. Clarify the problem and constraints

Confirm the operators (+, *), parentheses, and that numbers can be multi-digit. Ask about negative numbers, spaces, and whether division is included. State that you'll aim for O(n) time and O(n) space.

2. Design the stack-based algorithm

Use a stack to store the result so far and the sign before a parenthesis. Maintain a current number and a current multiplication segment. When encountering '+', add the segment to the result and start a new segment. When encountering '(', push the current result and sign, then reset. When encountering ')', finalize the current segment, apply the sign, add to the result, then pop and combine with the previous result.

3. Walk through an example

Trace the algorithm on a nested expression like '2*(3+(4*5+6))' to show how the stack and segments are updated. Highlight how multiplication is handled immediately and addition is deferred.

4. Analyze complexity and edge cases

State that each character is processed once, so time is O(n). Space is O(n) in the worst case due to stack depth for nested parentheses. Discuss edge cases: empty string, single number, negative numbers, multiple digits, and unbalanced parentheses.

Key Points to Mention

  • Operator precedence: multiplication is evaluated before addition, so maintain a current multiplication segment.
  • Stack usage: push the current result and sign when entering a parenthesis, pop and combine when exiting.
  • Handling nested parentheses: the stack naturally supports arbitrary nesting by saving and restoring context.
  • Time complexity O(n) because each character is processed once; space complexity O(n) for the stack in worst-case nesting.
  • Edge cases: negative numbers, multi-digit numbers, empty input, and unbalanced parentheses.
  • Alternative approaches: two-pass with stack or recursive descent, but iterative stack is optimal for O(n) time and space.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.